Sum of Normal Random Variable

Discussion in 'CT3' started by corpact90, Mar 18, 2016.

  1. corpact90

    corpact90 Member

    Hi everyone, I have a doubt regarding normal random variables.
    I know it could really easy to solve but still I can't figure this out.
    So, if you have to sum to gaussian variable as follows:
    aN(0,1)+bN(0,1)
    where a and b are constant we can proceed as follows:
    aN(0,1)+bN(0,1)=N(0,a^2)+N(0,b^2)=N(0,a^2+b^2)

    why is this different from this:
    aN(0,1)+bN(0,1)=(a+b)N(0,1)=N(0,(a+b)^2)

    which one of the two approaches is correct?

    Thanks
     
  2. djjain80

    djjain80 Member

    1st approach is correct..coz N(0'1) is a distribution variable but in second approach its been assumed as algebric variable...
    so its not sum of distribution but its sum of variable of a distribution that has algebric properties
     
  3. John Lee

    John Lee ActEd Tutor Staff Member

    Since \(a^2 + b^2 \neq (a + b)^2 \)
     

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